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UPSC CDS I 2025 Mathematics Paper

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Question : 28 of 100

Marks: +1, -0

If

\cot^{2}\theta - 3\sqrt{3} \cot\theta + 6 = 0

\frac{\pi}{6} < \theta < \frac{\pi}{2}

, then what is the value of

\sin\theta + \cos 2\theta

?

0
1
$\sqrt{3}$
$1+\sqrt{2}$

Solution:

Concept:

Solve a quadratic in cot θ by substitution, then evaluate the trigonometric expression.

Explanation:

Let

x = \cot \theta

. The equation becomes

x^2 - 3\sqrt{3}x + 6 = 0

.

Factor it as

(x - \sqrt{3})(x - 2\sqrt{3}) = 0

.

So

x = \sqrt{3}

or

x = 2\sqrt{3}

.

Since

\frac{\pi}{6} < \theta < \frac{\pi}{2}

, cot θ is positive. The value

\sqrt{3}

gives

\theta = 30^\circ = \frac{\pi}{6}

.

Now compute

\sin \theta + \cos 2\theta

with

\theta = 30^\circ

:

\sin 30^\circ = \frac12

and

\cos 60^\circ = \frac12

.

Their sum is

\frac12 + \frac12 = 1

.

Answer:

B. 1

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